3.347 \(\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^4} \, dx\)

Optimal. Leaf size=27 \[ -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \]

[Out]

-1/3*F^(a+b/(d*x+c)^3)/b/d/ln(F)

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Rubi [A]  time = 0.04, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2209} \[ -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^3)/(c + d*x)^4,x]

[Out]

-F^(a + b/(c + d*x)^3)/(3*b*d*Log[F])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^4} \, dx &=-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^4,x]

[Out]

-1/3*F^(a + b/(c + d*x)^3)/(b*d*Log[F])

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fricas [B]  time = 0.42, size = 77, normalized size = 2.85 \[ -\frac {F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/3*F^((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(b*d*
log(F))

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giac [B]  time = 0.48, size = 77, normalized size = 2.85 \[ -\frac {F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="giac")

[Out]

-1/3*F^((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(b*d*
log(F))

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maple [A]  time = 0.00, size = 26, normalized size = 0.96 \[ -\frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{3 b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^3*b)/(d*x+c)^4,x)

[Out]

-1/3*F^(a+1/(d*x+c)^3*b)/b/d/ln(F)

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maxima [A]  time = 0.61, size = 25, normalized size = 0.93 \[ -\frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/3*F^(a + b/(d*x + c)^3)/(b*d*log(F))

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mupad [B]  time = 3.64, size = 48, normalized size = 1.78 \[ -\frac {F^a\,F^{\frac {b}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3}}}{3\,b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^3)/(c + d*x)^4,x)

[Out]

-(F^a*F^(b/(c^3 + d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x)))/(3*b*d*log(F))

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sympy [A]  time = 0.40, size = 66, normalized size = 2.44 \[ \begin {cases} - \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}}}{3 b d \log {\relax (F )}} & \text {for}\: 3 b d \log {\relax (F )} \neq 0 \\- \frac {1}{3 c^{3} d + 9 c^{2} d^{2} x + 9 c d^{3} x^{2} + 3 d^{4} x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**4,x)

[Out]

Piecewise((-F**(a + b/(c + d*x)**3)/(3*b*d*log(F)), Ne(3*b*d*log(F), 0)), (-1/(3*c**3*d + 9*c**2*d**2*x + 9*c*
d**3*x**2 + 3*d**4*x**3), True))

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